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$N{m^{ - 1}}$ is the SI unit of
(A) Viscosity
(B) Spring constant
(C) Surface tension
(D) Both B and C

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Last updated date: 19th Jul 2024
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Answer
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Hint: They have provided a SI unit. We have to find the correct option that has $N{m^{ - 1}}$ as their unit. Newton per meter is used as the SI unit of various physical terms. Newton is the standard SI unit of force and meter is the standard SI unit of length.

Complete step by step answer
Viscosity: Viscosity is a measure of resistance of a fluid that deform at given rate. It is the opposition to flow. It corresponds to the concept of thickness. Example, castor oil has a higher viscosity than water. The reciprocal of the viscosity is known as fluidity, the measure of the ease of flow. This can be confusing when we deal with fluids of different densities.
Viscosity are of two types, Dynamic Viscosity or Absolute Viscosity and Kinematic viscosity. The SI unit of Dynamic Viscosity is pascal second, ${\text{pas s}}$. The SI unit of Kinematic viscosity is ${m^2}{s^{ - 1}}$
Surface tension: The tendency of surfaces to shrink into the possible minimum surface area. The surface tension is the force per unit length of the surface. Surface area of an object is given by
$ \Rightarrow \gamma = \dfrac{F}{l}$
$\gamma $ is the surface tension
F is the force
L is the unit length
The SI unit of surface tension is $N{m^{ - 1}}$. We can see that from the surface tension formula. As the formula says that surface tension is force by length whose SI units are newton and meter.
Spring constant: the spring constant is derived from Hooke's law. The Hooke's law says that the force required to stretch an elastic object (in case of a spring) is directly proportional to the extension of the spring.
$ \Rightarrow F = - kx$
The spring constant is a proportional constant. It is denoted by k. it is the measure of the spring's stiffness. i.e. when a spring is stretched or compressed its length changes by a quantity from its equilibrium length. The measure of the ability to change its elasticity is known as its stiffness.
$ \Rightarrow k = - \dfrac{F}{x}$
F is the force
K is the spring constant
X is the change in length
The SI unit of spring constant is $N{m^{ - 1}}$. Since spring constant is given by force per change in length
From above discussion we can see that surface tension and spring constant has $N{m^{ - 1}}$ as their SI unit

Hence the correct answer is option (D) both B and C

Note: The SI unit of Dynamic Viscosity is pascal second, ${\text{pas s}}$ which is equivalent to newton seconds per square of meter, ${\text{N s }}{{\text{m}}^{{\text{ - 2}}}}$ which is also equivalent to kilogram per meter per second, ${\text{Kg }}{{\text{m}}^{{\text{ - 1}}}}{\text{ }}{{\text{s}}^{{\text{ - 1}}}}$.